some quantitative characterizations of certain symplectic ... · some quantitative...

International Electronic Journal of Algebra

Volume 16 (2014) 32-52

SOME QUANTITATIVE CHARACTERIZATIONS OF CERTAIN

SYMPLECTIC GROUPS OVER THE BINARY FIELD

M. Akbari and A. R. Moghaddamfar

Received: 2 December 2013; Revised: 1 April 2014

Communicated by Arturo Magidin

Abstract. Given a finite group G, denote by D(G) the degree pattern of G

and by OC(G) the set of all order components of G. Denote by hOD(G) (resp.

hOC(G)) the number of isomorphism classes of finite groups H satisfying condi-

tions |H| = |G| and D(H) = D(G) (resp. OC(H) = OC(G)). A finite group G

is called OD-characterizable (resp. OC-characterizable) if hOD(G) = 1 (resp.

hOC(G) = 1). Let C = Cp(2) be a symplectic group over the binary field, for

which 2p − 1 > 7 is a Mersenne prime. The aim of this article is to prove that

hOD(C) = 1 = hOC(C).

Mathematics Subject Classification 2010: 20D05, 20D06, 20D08

Keywords: Spectrum of a group, prime graph, degree pattern, order compo-

nent, symplectic group Cn(q), OD(OC)-characterizability of a finite group

1. Introduction

Only finite groups will be considered. Let G be a group, π(G) the set of all prime

divisors of its order and ω(G) be the spectrum of G, that is the set of its element

orders. The prime graph GK(G) (or Gruenberg-Kegel graph) of G is a simple graph

whose vertex set is π(G) and two distinct vertices p and q are joined by an edge if

and only if pq ∈ ω(G). Let t(G) be the number of connected components of GK(G).

The vertex set of the ith connected component of GK(G) is denoted by πi(G) for

each i = 1, 2, . . . , t(G). In the case when 2 ∈ π(G), we assume that 2 ∈ π1(G). The

classification of finite simple groups with disconnected prime graph was obtained by

Williams [13] and Kondratev [4]. Recall that a clique in a graph is a set of pairwise

adjacent vertices. Note that for all non-abelian simple groups S with disconnected

prime graph, all connected components πi(S) for 2 6 i 6 t(S) are cliques, for

instance, see [13]. The degree degG(p) of a vertex p ∈ π(G) in GK(G) is the

number of edges incident on p. If π(G) = {p1, p2, . . . , ph} with p1 < p2 < · · · < ph,

then we define

D(G) =(

degG(p1),degG(p2), . . . ,degG(ph)),

CERTAIN SYMPLECTIC GROUPS OVER THE BINARY FIELD 33

which is called the degree pattern of G. Given a group G, denote by hOD(G) the

number of isomorphism classes of groups with the same order and degree pattern

as G. All finite groups, in terms of the function hOD(·), are classified as follows:

Definition 1.1. A group G is called k-fold OD-characterizable if hOD(G) = k.

Usually, a 1-fold OD-characterizable group is simply called OD-characterizable.

There are scattered results in the literature showing that certain simple groups

are k-fold OD-characterizable for k ∈ {1, 2}. The most recent version of the list of

such simple groups is presented in [8, Tables 2 and 3]. Until now, no examples of

simple groups S with hOD(S) > 3 were known. Therefore, we posed the following

question:

Problem 1.2. Is there a non-abelian simple group S with hOD(S) > 3?

In this article, we focus our attention on the symplectic groups Cp(2) ∼= S2p(2),

where p is an odd prime. Recall that C2(2) is not a simple group, in fact, the

derived subgroup C2(2)′ is a simple group which is isomorphic with A6∼= L2(9).

In addition, we recall that B2(3) ∼= 2A4(22), Bn(2m) ∼= Cn(2m) and B2(q) ∼= C2(q)

(see [2]). Previously, it was determined the values of hOD(·) for some symplectic

and orthogonal groups (see [1,6,9]). In the table below, π(n) is the set of all prime

divisors of n, where n is a natural number.

G Restrictions on G hOD(G) Refs.

B3(4) ∼= C3(4) 1 [6]

B2(q) ∼= C2(q) |π( q2+1(2,q−1) )| = 1 1 [1]

B2m(q) ∼= C2m(q) |π( q2+1(2,q−1) )| = 1, q is even 1 [1]

B3(5), C3(5), 2 [1]

Bn(q), Cn(q), n = 2m > 2, |π( qn+12 )| = 1, 2 [1]

q is an odd prime power

Bp(3), Cp(3), |π( 3p−12 )| = 1, p is an odd prime, 2 [1,9]

Given a group G, the order of G can be expressed as a product of some coprime

natural numbers mi(G), i = 1, 2, . . . , t(G), with π(mi(G)) = πi(G). The numbers

m1(G),m2(G), . . . ,mt(G)(G) are called the order components of G. We set

OC(G) = {m1(G),m2(G), . . . ,mt(G)(G)}.

34 M. AKBARI AND A. R. MOGHADDAMFAR

In a similar manner, we define hOC(G) as the number of isomorphism classes of

finite groups with the same set OC(G) of order components. Again, in terms of

function hOC(·), the groups G are classified as follows:

Definition 1.3. A finite group G is called k-fold OC-characterizable if hOC(G) = k.

In the case when k = 1 the group G is simply called OC-characterizable.

A Mersenne prime is a prime that can be written as 2p − 1 for some prime p.

The purpose of this article is to prove the following theorem.

Main Theorem. Let C = Cp(2) be the symplectic group over the binary field, for

which 2p − 1 > 7 is a Mersenne prime. Then hOD(C) = 1 = hOC(C).

It is worth noting that the values of functions hOD(·) and hOC(·) may be different.

For instance, supposeM ∈ {B3(5), C3(5)}. By [13], the prime graph associated with

M is connected and so OC(M) = {|M |} = {29 · 34 · 59 · 7 · 13 · 31}. On the other

hand, it is easy to see that the prime graph associated with a nilpotent group is

always a clique, hence, we have

hOC(M) > νnil(|M |) > νa(|M |) = Par(9)2 · Par(4) = 302 × 5 = 4500,

where νnil(n) (resp. νa(n)) signifies the number of non-isomorphic nilpotent (resp.

abelian) groups of order n and Par(n) denotes the number of partitions of n. How-

ever, by Theorem 1.3 in [1], we know that hOD(M) = 2.

2. Preliminaries

If a is a natural number, r is an odd prime and (r, a) = 1, then by e(r, a)

we denote the multiplicative order of a modulo r, that is the minimal natural

number n with an ≡ 1 (mod r). If a is odd, we put e(2, a) = 1 if a ≡ 1 (mod 4),

and e(2, a) = 2 if a ≡ −1 (mod 4). The following lemma is a consequence of

Zsigmondy’s Theorem (see [14]).

Lemma 2.1. Let a > 1 be an integer. Then for every natural number n there exists

a prime r with e(r, a) = n except for the cases (n, a) ∈ {(1, 2), (1, 3), (6, 2)}.

A prime r with e(r, a) = n is called a primitive prime divisor of an − 1. By

Lemma 2.1, such a prime exists except for the cases mentioned in the lemma. We

denote by ppd(an − 1) the set of all primitive prime divisors of an − 1. By our

definition, we have π(a − 1) = ppd(a − 1) but for the following sole exception,

namely, 2 /∈ ppd(a− 1) if e(2, a) = 2. In this case, we assume that 2 ∈ ppd(a2− 1).


From the definition it is easy to conclude that: Let p > 2 be an integer. Then

π(ap − 1) = ppd(ap − 1) if and only if p is a prime.

In the following results, we will consider the function η : N→ N, which is defined

as follows

η(m) =

{m if m ≡ 1 (mod 2),

m/2 if m ≡ 0 (mod 2).

Lemma 2.2. ([11]) Let M be one of the simple groups of Lie type Bn(q) or Cn(q)

over a field of characteristic p, and let r ∈ π(M) \ {p} and r ∈ ppd(qk − 1). Then

r and p are non-adjacent if and only if η(k) > n− 1.

Lemma 2.3. ([12]) Let M be one of the simple groups of Lie type Bn(q) or Cn(q)

over a field of characteristic p. Let r, s be odd primes with r, s ∈ π(M) \ {p}.Suppose that r ∈ ppd(qk − 1), s ∈ ppd(ql − 1) and 1 6 η(k) 6 η(l). Then r and

s are non-adjacent if and only if η(k) + η(l) > n and l/k is not an odd natural

number.

Using Lemmas 2.2 and 2.3, we conclude that the prime graphs GK(Bn(q)) and

GK(Cn(q)) coincide (see also [11, Proposition 7.5]), and hence

D(Bn(q)) = D(Cn(q)).

Corollary 2.4. Let p > 3 be a prime and C = Cp(2). Then degC(3) = |π1(C)|−1.

Proof. Recall that, by [4], we have

π1(C) = π

(2(2p + 1)

p−1∏i=1

(22i − 1)

)and π2(C) = π(2p − 1).

Now, it follows from Lemma 2.3 that all primitive prime divisors of 2p − 1 (and so

all primes in π(2p − 1)) are non-adjacent to 3. On the other hand, by Lemmas 2.2

and 2.3, we deduce that degC(3) = |π1(C)| − 1, as desired. �

The following lemma is crucial to the study of characterizability of symplectic

groups Cp(2) by order components.

Lemma 2.5. ([3]) Let G be a group whose prime graph has more than one com-

ponent. If H is a normal πk(G)-subgroup of G, then |H| − 1 is divisible by mi(G),

for all i 6= k.

A group G is called 2-Frobenius if there exists a normal series 1 E H E K E G

of G such that H is the Frobenius kernel of K and K/H is the Frobenius kernel of

G/H.


Lemma 2.6. ([7]) Let S be a simple group with disconnected prime graph GK(S),

except U4(2) and U5(2). If G is a finite group with OC(G) = OC(S), then G is

neither a Frobenius group nor a 2-Frobenius group.

Lemma 2.7. ([13]) Let G be a group with t(G) > 2. Then one of the following

hold:

(1) G is a Frobenius group;

(2) G is a 2-Frobenius group; or

(3) G has a normal series 1EH CK EG such that H is a nilpotent π1-group,

K/H is a non-abelian simple group, G/K is a π1-group, |G/K| divides

|Out(K/H)| and any odd order component of G is equal to one of the odd

order components of K/H.

Lemma 2.8. ([5]) The only solution of the equation pm − qn = 1, where p, q are

primes and m,n > 1 are integers, is (p, q,m, n) = (3, 2, 2, 3).

Given a natural number B and a prime number t, we denote by Bt the t-part of

B, that is the largest power of t dividing B.

Lemma 2.9. ([10]) Let B = (22 − 1)(24 − 1) · · · (22n − 1). If t is a prime divisor

of B, then Bt < 23n. Furthermore, if t > 5 then Bt < 22n.

3. Proof of the main theorem

Throughout this section, we will assume that 2p − 1 > 7 is a Mersenne prime

and C = Cp(2). Suppose that G is a group with the same order and degree pattern

as C, that is

|G| = |C| = 2p2p∏i=1

(22i − 1) and D(G) = D(C).

Note that, according to the results summarized in [4], we have t(C) = 2, and

π1(C) = π

(2(2p + 1)

p−1∏i=1

(22i − 1)

)and π2(C) = {2p − 1}.

By our hypothesis, it is easy to see that

π2(G) = π2(C) = {2p − 1} and π(G) = π(C) = π1(C) ∪ {2p − 1}.

First of all, we notice that 2p − 1 is the largest prime in π(G) = π(C). Moreover,

it follows from Corollary 2.4 that

degG(3) = degC(3) = |π1(C)| − 1,


and this forces π1(G) = π1(C), and so t(G) = 2. Hence, we have

OC(G) = OC(C) =

{2p

2

(2p + 1)

p−1∏i=1

(22i − 1), 2p − 1

},

and from Lemma 2.6, the group G is neither a Frobenius group nor a 2-Frobenius

group. Finally, Lemma 2.7, reduces the problem to the study of the simple groups.

Indeed, by Lemma 2.7, there is a normal series 1EH CK EG of G such that:

(1) H is a nilpotent π1(G)-group, K/H is a non-abelian simple group and

G/K is a π1(G)-group. Moreover, we have K/H 6 G/H 6 Aut(K/H),

and t(K/H) > t(G) > 2,

(2) 2p − 1 is the only odd order component of G which is equal to one of those

of the quotient K/H,

(3) |G/K| divides |Out(K/H)|.

For odd order components of K/H see [4,13]. Now, we will continue the proof

step by step.

Step 3.1. K/H � 2A3(2), 2F4(2)′, 2A5(2), E7(2), E7(3), A2(4), 2E6(2) nor one of

the sporadic simple groups.

Note that either the odd order components of above groups are not equal to a

Mersenne prime 2p − 1 > 7 or their orders do not divide the order of G.

In the following, An denotes the alternating group on n letters.

Step 3.2. K/H � An, where n and n− 2 are both prime numbers.

In this case, it follows that n = 2p − 1. Now, simple computations show that

|An|2 =

(n!

2

)2

= 2

([n2

]+[

n22

]+···)−1 = 22

p−p−2.

If p > 5, then 2p − p − 2 > p2 and hence the 2-part of |An| does not divide the

2-part of |G|, i.e. 2p2

, which is a contradiction. In the case when p = 5, then n = 31

and |K/H| = (31!)/2, which does not divide |G| = |C5(2)| = 225 ·36 ·52 ·7 ·11 ·17 ·31,

which is again a contradiction.

Step 3.3. K/H � An, where n = q, q + 1, or q + 2 (q is a prime), and one of n,

n− 2 is not prime.

Here, q is the only odd order component of K/H, and so q = 2p − 1. We now

consider the alternating group Aq which is a subgroup of K/H ∼= An. Similar

arguments as those in the previous step, on the subgroup Aq instead of An, lead us

a contradiction.


Step 3.4. K/H is isomorphic to neither 2E6(q), q > 2, nor E6(q).

We deal with 2E6(q), q > 2, the proof for E6(q) being quite similar. Suppose

that K/H ∼= 2E6(q). First of all, we recall that

|2E6(q)| = 1

(3, q + 1)q36(q12 − 1)(q9 + 1)(q8 − 1)(q6 − 1)(q5 + 1)(q2 − 1).

Considering the only odd order component of 2E6(q), that is (q6−q3+1)/(3, q+1),

we must have (q6 − q3 + 1)/(3, q + 1) = 2p − 1, which implies that q9 > 2p, or

equivalently q36 > 24p. Let q = rf . If r is an odd prime, then from Lemma 2.9, we

get

q36 = r36f = |K/H|r 6 |G|r < 23p,

which is a contradiction. Therefore we may assume that r = 2. In this case, we

have

(26f − 23f + 1)/(3, 2f + 1) = 2p − 1.

Now, if (3, 2f + 1) = 1, then we obtain 23f (23f − 1) = 2(2p−1 − 1), from which we

deduce that 3f = 1, a contradiction. In the case where (3, 2f + 1) = 3, an easy

calculation shows that

23f (23f − 1) = 22(3 · 2p−2 − 1),

and so 3f = 2, which is again a contradiction.

Step 3.5. K/H � F4(q), where q is an odd prime power.

We remark that q4−q2 +1 is the only odd order component of F4(q), and clearly

this forces q4 − q2 + 1 = 2p − 1. Then q2(q2 − 1) = 2(2p−1 − 1), which shows that

2(2p−1 − 1) is divisible by 4, a contradiction.

Step 3.6. K/H � 2B2(q), where q = 22m+1 > 2.

Recall that |2B2(q)| = q2(q2 +1)(q−1) and the odd order components of 2B2(q)

are:

q − 1, q −√

2q + 1, q +√

2q + 1.

If q − 1 = 2p − 1, then q = 2p. Now, we consider the primitive prime divisor

r ∈ ppd(24p − 1). Clearly r ∈ π(22p + 1), and so r ∈ π(2B2(q)) ⊆ π(G). This is a

contradiction.

In the case when

q −√

2q + 1 = 2p − 1 (resp. q +√

2q + 1 = 2p − 1),


by simple computations we obtain

2m+1(2m − 1) = 2(2p−1 − 1) (resp. 2m+1(2m + 1) = 2(2p−1 − 1)),

a contradiction.

Step 3.7. K/H � E8(q), where q ≡ 2, 3 (mod 5).

The odd order components of E8(q) in this case are

q8 − q4 + 1,q10 + q5 + 1

q2 + q + 1,

q10 − q5 + 1

q2 − q + 1.

If q8 − q4 + 1 = 2p − 1, then we obtain q4(q − 1)(q + 1)(q2 + 1) = 2(2p−1 − 1).

However, the left hand side is divisible by 16, while the right hand side is not

divisible by 4, which is impossible.

If (q10 + q5 + 1)/(q2 + q + 1) = 2p − 1, then after subtracting 1 from both sides

of this equation and some simple computations, we obtain

q(q − 1)(q + 1)(q2 + 1)(q3 − q2 + 1) = 2(2p−1 − 1).

Now, if q is odd, then the left hand side is divisible by 16, a contradiction. Moreover,

if q is even, then it follows that q = 2, and if this is substituted in above equation

we get 76 = 2p−1, a contradiction.

The case (q10− q5 + 1)/(q2− q+ 1) = 2p− 1 is quite similar to the previous case

and it is omitted.

Step 3.8. K/H � E8(q), where q ≡ 0, 1, 4 (mod 5).

The odd order components of E8(q) in this case are

q10 + 1

q2 + 1, q8 − q4 + 1,

q10 + q5 + 1

q2 + q + 1,

q10 − q5 + 1

q2 − q + 1.

Consider the first case. Let (q10 + 1)/(q2 + 1) = 2p− 1. Subtracting 1 from both

sides of this equality, we get

q2(q2 − 1)(q4 + 1) = 2(2p−1 − 1),

which implies 2(2p−1 − 1) is divisible by 4, a contradiction.

Similarly, if q8−q4 +1 = 2p−1, we obtain q4(q−1)(q+1)(q2 +1) = 2(2p−1−1),

which shows that 2(2p−1− 1) is divisible by 16, a contradiction. Similar arguments

work if (q10 + q5 + 1)/(q2 + q + 1) = 2p − 1 or (q10 − q5 + 1)/(q2 − q + 1) = 2p − 1,

and we omit the details.

Step 3.9. K/H � 2F 4(q), where q = 22m+1 > 2.


The odd order components of 2F4(q) are:

q2 +√

2q3 + q +√

2q + 1 and q2 −√

2q3 + q −√

2q + 1.

Therefore, we have

q2 +√

2q3 + q +√

2q + 1 = 2p − 1 or q2 −√

2q3 + q −√

2q + 1 = 2p − 1.

However, if 22m+1 is substituted in these equations we obtain

2m+1(23m+1 ± 22m+1 + 2m ± 1) = 2(2p−1 − 1),

which is a contradiction.

Step 3.10. K/H � F4(q), where q = 2m.

The odd order components of F4(q) are q4+1 and q4−q2+1, hence q4+1 = 2p−1

or q4 − q2 + 1 = 2p − 1. Now, it is easy to see that in both cases, 22m divides

2(2p−1 − 1), a contradiction.

Step 3.11. K/H � 2G2(q), where q = 32m+1 > 3.

The odd order components of 2G2(q) are q +√

3q + 1 and q −√

3q + 1. If

q −√

3q + 1 = 2p − 1, then q3 > 23p, while Lemma 2.9 shows that q3 < 23p, which

is a contradiction. If q +√

3q + 1 = 2p − 1, then

2p − 2 = 2(2(p−1)/2 − 1)(2(p−1)/2 + 1) = 3m+1(3m + 1). (1)

First of all, we recall that (2(p−1)/2 − 1, 2(p−1)/2 + 1) = 1. Now we consider two

cases separately:

(i) If 3m+1 divides 2(p−1)/2 − 1, then

3m + 1 < 3m+1 6 2(p−1)/2 − 1 < 2(p−1)/2 + 1.

Hence, we obtain

3m+1(3m + 1) < 2(2(p−1)/2 − 1)(2(p−1)/2 + 1),

a contradiction.

(ii) If 3m+1 divides 2(p−1)/2+1, then 2(p−1)/2+1 = k ·3m+1 where k is a natural

number. Now, from Eq.( 1), it follows that

2k(2(p−1)/2 − 1) = 3m + 1,

and consequently 3m > 2(p+1)/2 − 1. Therefore we have

2(p+1)/2 − 1 6 3m < 3m+1 6 2(p−1)/2 + 1,

a contradiction.


Step 3.12. K/H � G2(q), where q = 3m.

Recall that the odd order components of G2(q) are q2 − q + 1 and q2 + q + 1. If

q2 − q + 1 = 2p − 1 then q6 > 23p, while one can follow from Lemma 2.9 that

q6 < 23p, which is a contradiction. If q2 +q+1 = 2p−1, then q(q+1) ≡ 2 (mod 4),

which forces m is even. But then, it is obvious that 2p− 2 = q(q+ 1) ≡ 2 (mod 8),

a contradiction.

Step 3.13. K/H � 2Dr(3), where r = 2m + 1 is a prime number and m > 1.

Recall that

|2Dr(3)| = 1

(4, 3r + 1)3r(r−1)(3r + 1)

r−1∏i=1

(32i − 1),

and the odd order components of 2Dr(3) are

(3r−1 + 1)/2 and (3r + 1)/4.

In the case when (3r−1 + 1)/2 = 2p − 1, adding 1 to both sides of this equality, we

obtain

3(3r−2 + 1) = 2p+1,

which is a contradiction. If (3r+1)/4 = 2p−1, then r > 5 because p > 5. Moreover,

on the one hand, from last equation we obtain 3r = 2p+2− 5 > 2p+1, which implies

that

3r(r−1) > 2(p+1)(r−1) > 24(p+1).

On the other hand, it follows from Lemma 2.9 that

3r(r−1) = |K/H|3 6 |G|3 < 23p,


Step 3.14. K/H � Bn(q), where n = 2m > 4 and q = rf is an odd prime power.

Note that

|Bn(q)| = 1

(2, q − 1)qn

2n∏i=1

(q2i − 1),

and the only odd order component of Bn(q) is (qn + 1)/2. If (qn + 1)/2 = 2p − 1,

then qn = 2p+1 − 3 > 2p and clearly q is not divisible by 2 and 3. Since p > 5 and

n > 4, it is easy to see that

qn2

> q3n > 23p > 22p.

On the other hand, by Lemma 2.9, we obtain

qn2

= |K/H|r 6 |G|r < 22p,



Step 3.15. K/H � Br(3).

The only odd order component of Br(3) is (3r − 1)/2. If (3r − 1)/2 = 2p − 1,

then 2p+1 − 3r = 1. However, this equation has no solution by Lemma 2.8, which

is impossible.

Step 3.16. K/H � 3D4(q).

We recall that q4 − q2 + 1 is the only odd order component of 3D4(q), and so

q4−q2+1 = 2p−1. But then, q2(q2−1) = 2(2p−1−1), which shows that 2(2p−1−1)

is divisible by 4, a contradiction.

Step 3.17. K/H � G2(q), where 2 < q ≡ ±1 (mod 3).

In this case, the odd order components of G2(q) are q2 + q + 1 and q2 − q + 1.

Let q = rf . If q2 + q + 1 = 2p − 1, then q(q + 1) = 2(2p−1 − 1), which shows that

q > 2 is not a power of 2. Moreover, since q − 1 > 2, we obtain

q3 − 1 = (q − 1)(q2 + q + 1) > 2(2p − 1),

and so q3 > 2p+1 − 1 > 2p, which yields that q6 > 22p. However, since

|G2(q)| = q6(q2 − 1)(q6 − 1),

from Lemma 2.9, we conclude that

q6 = |K/H|r 6 |G|r < 22p,


The case when q2 − q + 1 = 2p − 1 is similar and left to the reader.

Step 3.18. K/H � 2Dn(3), where n = 2m + 1 which is not a prime and m > 2.

The odd order component of 2Dn(3) is (3n−1 + 1)/2. If (3n−1 + 1)/2 = 2p − 1,

then 2p+1 = 3(3n−2 + 1), a contradiction.

Step 3.19. K/H � 2Dr(3), where r > 5 is a prime and r 6= 2m + 1.

Here, we have

|2Dr(3)| = 1

(4, 3r + 1)3r(r−1)(3r + 1)

r−1∏i=1

(32i − 1).


The only odd order component of 2Dr(3) is (3r + 1)/4, and so (3r + 1)/4 = 2p − 1.

An easy computation shows that 3r = 2p+2 − 5 > 2p+1. Moreover, we note that

r − 1 > 4, and so

3r(r−1) > 34r > 24(p+1).

On the other hand, by Lemma 2.9, we obtain

3r(r−1) = |K/H|3 6 |G|3 < 23p,


Step 3.20. K/H � 2Dn(2), where n = 2m + 1, m > 2.

The only odd order component of 2Dn(2) is 2n−1 + 1. Therefore, we obtain

2n−1 + 1 = 2p − 1, which is impossible.

Step 3.21. K/H � 2Dn(q), where n = 2m > 4 and q = rf .

Recall that

|2Dn(q)| = 1

(4, qn + 1)qn(n−1)(qn + 1)

n−1∏i=1

(q2i − 1),

and the only odd order component of 2Dn(q) is (qn + 1)/(2, q + 1). Therefore,

(qn+ 1)/(2, q+ 1) = 2p−1. Assume first that (2, q+ 1) = 1. In this case, we obtain

qn = 2(2p−1 − 1), a contradiction. Assume next that (2, q + 1) = 2. Again, using

simple calculations we obtain qn = 2p+1 − 3 > 2p and so q cannot be a power of 2.

Moreover, since n− 1 > 3, qn(n−1) > q3n > 23p. Now, Lemma 2.9 shows that

qn(n−1) = |K/H|r 6 |G|r < 23p,


Step 3.22. K/H � Dr+1(q), where q = 2, 3.

Since, the only odd order component of Dr+1(q) is (qr − 1)/(2, q − 1), we have

(qr − 1)/(2, q − 1) = 2p − 1. If (2, q − 1) = 1, then r = p and q = 2, and we have

|K/H| = |Dp+1(2)| = 1

(4, 2p+1 − 1)2p(p+1)(2p+1 − 1)

p∏i=1

(22i − 1),

this shows that |K/H|2 = 2p(p+1)/(4, 2p+1−1) does not divide |G|2 = 2p2

, which is

a contradiction. In the case when (2, q−1) = 2, we have the equation 2p+1−3r = 1,

which has no solution for p > 5, by Lemma 2.8. This is again a contradiction.

Step 3.23. K/H � Dr(q), where q = 2, 3, 5 and r > 5.


We recall that the only odd order component of Dr(q) is (qr − 1)/(q − 1). We

distinguish three cases separately.

(i) q = 2. In this case, we have 2r − 1 = 2p − 1, and so r = p and

|K/H| = |Dp(2)| = 2p(p−1)(2p − 1)

p−1∏i=1

(22i − 1).

Note that |Out(Dp(2))| = 2 and Dp(2) 6 G/H 6 Aut(Dp(2)). Now,

considering the order of groups, we get |H| = 2α(2p+1) where p−1 6 α 6 p.

Let r ∈ ppd(22p−1) and Q ∈ Sylr(H). Clearly r ∈ π(2p+1), Q is a normal

π1(G)-subgroup of G and |Q| divides 2p + 1. Now, from Lemma 2.5, it

follows that |Q| − 1 is divisible by m2(G) = 2p − 1, and so |Q| − 1 > 2p − 1

or equivalently |Q| > 2p. This forces |Q| = 2p+1. But then m2(G) = 2p−1

does not divide the value |Q| − 1 = 2p, which is a contradiction.

(ii) q = 3. In this case, from the equality (3r − 1)/2 = 2p − 1, we deduce that

2p+1 − 3r = 1. However, this equation has no solution when p > 5 by

Lemma 2.8, a contradiction.

(iii) q = 5. Here (5r − 1)/4 = 2p − 1, and so 5r = 2p+2 − 3 > 2p+1. As before,

since r − 1 > 4, we obtain 5r(r−1) > 54r > 24(p+1). On the other hand, by

Lemma 2.9, we have

5r(r−1) = |K/H|5 6 |G|5 < 22p,


Step 3.24. K/H � Cr(3).

The only odd order component of Cr(3) is (3r − 1)/2. Thus, if (3r − 1)/2 = 2p− 1,

then 2p+1 − 3r = 1. However, this equation has no solution by Lemma 2.8, which

is impossible.

Step 3.25. K/H � Cn(q), where n = 2m > 2.

Note that

|Cn(q)| = 1

(2, q − 1)qn

2n∏i=1

(q2i − 1),

and the only odd order component of Cn(q) is (qn + 1)/(2, q − 1). Therefore,

(qn + 1)/(2, q − 1) = 2p − 1. If (2, q − 1) = 1, then qn = 2(2p−1 − 1), which yields

that q = p = 2 and n = 1, a contradiction. If (2, q−1) = 2, then qn = 2p+1−3 > 2p,

which implies that q is not a power of 2 and 3. Let q = rf . When n > 4, it is easy

to see that

qn2

> q3n > 23p > 22(p+1).


But, from Lemma 2.9, we obtain

qn2

= |K/H|r 6 |G|r < 22p,

a contradiction. Assume now that n = 2. In this case, we have q2 = 2p+1 − 3, or

equivalently

(q − 1)(q + 1) = 22(2p−1 − 1).

However, the left hand side is divisible by 8, while the right hand side is divisible

by 4, a contradiction.

Step 3.26. K/H � A1(q), where q = 2m > 2.

The odd order components of A1(q) are q + 1 and q − 1. If q + 1 = 2p − 1, then

q = 2(2p−1 − 1), a contradiction. If q − 1 = 2p − 1, then q = 2p. Moreover, since

A1(q) 6 G/H 6 Aut(A1(q)), it is easy to see that the order of H is divisible by

(22−1)(24−1) · · · (22(p−1)−1). Let r ∈ ppd(22(p−1)−1) and Q ∈ Sylr(H). Clearly

Q is a normal π1(G)-subgroup of G and |Q| divides 2p−1 + 1. On the other hand,

from Lemma 2.5, |Q| − 1 is divisible by 2p − 1 which implies that |Q| > 2p. This is

a contradiction.

Step 3.27. K/H � A1(q), where 3 6 q ≡ ±1 (mod 4) and q = rf .

Assume first that 3 6 q ≡ 1 (mod 4). In this case, the odd order components of

A1(q) are (q + 1)/2 and q. If (q + 1)/2 = 2p − 1, then rf = q = 2p+1 − 3. First of

all, we claim that f is an odd number. Otherwise, we have

(r f/2 − 1)(r f/2 + 1) = 22(2p−1 − 1).

But then, the left hand side is divisible by 8, while the right hand side is divisible

by 4, which is a contradiction. Furthermore, by easy computations we observe that

|A1(q)| = 1

2q(q2 − 1) = 22(2p+1 − 3)(2p−1 − 1)(2p − 1).

On the other hand, we have |G/K| · |H| = |G|/|A1(q)|, from which we deduce that

|G/K|2 · |H|2 =|G|2|A1(q)|2

= 2p2−2.

But since |G/K| divides |Out(A1(q))| = 2f and f is odd, |G/K|2 is at most 2.

Hence, if S2 ∈ Syl2(H), then |S2| = 2p2−2 or |S2| = 2p

2−3. We notice that S2 is

a normal subgroup of G, because H is nilpotent. Now, it follows from Lemma 2.5

that 2p − 1 divides 2p2−2 − 1 or 2p

2−3 − 1, which is a contradiction. If q = 2p − 1,

we get a contradiction by Lemma 2.8.

Assume next that 3 6 q ≡ −1 (mod 4). In this case, the odd order components

of A1(q) are (q − 1)/2 and q. If (q − 1)/2 = 2p − 1, then 2p+1 − rf = 1. Noting


Lemma 2.8, we deduce that f = 1, and hence r = 2p+1 − 1 is a Mersenne prime,

which is a contradiction because p+ 1 is not a prime.

The case when q = 2p − 1 is similar to the previous paragraph.

Step 3.28. K/H � Ar(q), where (q − 1)∣∣(r + 1).

Recall that

|K/H| = |Ar(q)| =1

(r + 1, q − 1)qr(r+1)/2

r+1∏i=2

(qi − 1).

The only odd order component of Ar(q) is (qr − 1)/(q − 1), and so

(qr − 1)/(q − 1) = 2p − 1.

As a simple observation we see that qr − 1 > (qr − 1)/(q − 1) = 2p − 1 and so

qr > 2p. Let q = tf , where t is a prime number and f is a natural number.

(i) Suppose first that r > 7. Then qr(r+1)/2 > q3(r+1) > 23q3r > 23(p+1). Now,

if t is an odd prime, then by Lemma 2.9 we obtain

qr(r+1)/2 = |K/H|t 6 |G|t < 23p,

which is a contradiction. Therefore, we may assume that t = 2. In this

case, we have

(2fr − 1)/(2f − 1) = 2p − 1,

from which one can deduce that f = 1 and r = p. Thus

|G/K| · |H| =2p

2 ∏pi=1(22i − 1)

2p(p+1)

2

∏p+1i=2 (2i − 1)

.

Since |G/K| divides |Out(K/H)| = |Out(Ap(2))| = 2, we conclude that |H|is divisible by 2p + 1. Let s ∈ ppd(22p − 1) ⊆ π(2p + 1) and Q ∈ Syls(H).

Clearly |Q|∣∣2p + 1. Since H is a normal π1(G)-subgroup of G which is

nilpotent, Q is also a normal π1(G)-subgroup of G. Now, by Lemma 2.5,

m2(G) = 2p−1 divides |Q|−1, and so |Q| > 2p. But, this forces |Q| = 2p+1.

However, this contradicts the fact that m2(G)||Q| − 1.

(ii) Suppose next that r = 5. If q is even, then from (q5 − 1)/(q − 1) = 2p − 1,

we obtain q(q3 + q2 + q + 1) = 2(2p−1 − 1), which implies that q = 2 and

r = p = 5. Therefore, by easy calculations we see that

|G/K| · |H| =210∏5i=1(2i + 1)

26 − 1,

which is not a natural number, a contradiction. If q is odd, then we get

q(q + 1)(q2 + 1) = q4 + q3 + q2 + q = 2p − 2,


however q(q+ 1)(q2 + 1) ≡ 0 (mod 4), while 2p − 2 ≡ 2 (mod 4), a contra-

diction.

(iii) Finally suppose that r = 3. Then q(q + 1) = 2(2p−1 − 1). First of all, we

note that q is not even, otherwise p = 3, which is impossible. In addition,

we have

q(q + 1) = 2(2(p−1)/2 − 1)(2(p−1)/2 + 1). (2)

Now we consider two cases separately:

(a) If q divides 2(p−1)/2 − 1, then

q 6 2(p−1)/2 − 1, q + 1 < 2(p−1)/2 + 1.

Hence, we obtain

q(q + 1) < 2(2(p−1)/2 − 1)(2(p−1)/2 + 1),

a contradiction.

(b) If q divides 2(p−1)/2 + 1, then 2(p−1)/2 + 1 = kq for some natural

number k. Now from Eq.( 2), it follows that

2k(2(p−1)/2 − 1) = q + 1.

If k = 1, then p = q = 5. Hence 13 ∈ π(K/H) = π(A3(5)), however

13 /∈ π(G) = π(C5(2)), a contradiction. Thus, k > 2 and we obtain

2(2(p+1)/2 − 2)− 1 6 q < q + 1 6 kq = 2(p−1)/2 + 1,


Step 3.29. K/H � Ar−1(q), where (r, q) 6= (3, 2), (3, 4).

Again, we recall that

|K/H| = |Ar−1(q)| = 1

(r, q − 1)qr(r−1)/2

r∏i=2

(qi − 1),

and the only odd order component of Ar−1(q) is (qr − 1)/(q − 1)(r, q − 1). Hence,

we must have

(qr − 1)/(q − 1)(r, q − 1) = 2p − 1,

which implies that

qr − 1 > (qr − 1)/(q − 1)(r, q − 1) = 2p − 1,

or equivalently qr > 2p. Let q = tf , where t is a prime and f is a natural number.

In what follows, we consider several cases separately.


(i) r > 7. In this case, we obtain

qr(r−1)/2 > q3r > 23p,

and Lemma 2.9 implies that t = 2. Now, Lemma 2.1 shows that q = 2 and

r = p, and hence we obtain

|G/K| · |H| =2p

2 ∏pi=1(22i − 1)

2(p2)∏pi=2(2i − 1)

= 2p(p+1)

2

p∏i=1

(2i + 1).

On the other hand, |G/K| divides |Out(K/H)| = 2. From this we deduce

that |H| is divisible by 2p + 1. Let s ∈ ppd(22p − 1) ⊆ π(2p + 1) and

Q ∈ Syls(H). Evidently Q is a normal subgroup of G and |Q| divides

2p+1. Now, it follows from Lemma 2.5 that m2(G) = 2p−1∣∣|Q|−1, which

is impossible.

(ii) r = 5. Assume first that (5, q − 1) = 1. In this case, we have

q5 − 1

q − 1= q4 + q3 + q2 + q + 1 = 2p − 1,

or equivalently

q(q + 1)(q2 + 1) = 2(2p−1 − 1). (3)

If q is even, then we conclude that q = 2 and r = p = 5, and the proof

is quite similar as (i). If q is odd, then the left-hand side of Eq.( 3) is

congruent to 0 (mod 4), while the right-hand side of Eq.( 3) is congruent

to 2 (mod 4), a contradiction.

Assume next that (5, q − 1) = 5. In this case, we have

q4 + q3 + q2 + q + 1 = 5(2p − 1),

or equivalently

(q − 1)(q3 + 2q2 + 3q + 4) = 10(2p−1 − 1).

In the case when q is even, one can easily deduce that q = 2, and so

13 = 5(2p−1 − 1), a contradiction. Moreover, if q is odd, then from the

equality q(q + 1)(q2 + 1) = 5 · 2p − 6 it is easily seen that the left-hand

side of this equation is congruent to 0 (mod 4), while the right-hand side

is congruent to 2 (mod 4), a contradiction.

(iii) r = 3. In this case, we have (q3 − 1)/(q − 1)(3, q − 1) = 2p − 1. First of

all, if q is even, then we obtain p = 3, which is not the case. Thus, we can

assume that q is odd.


If (3, q − 1) = 1, then

q(q + 1) = 2(2(p−1)/2 − 1)(2(p−1)/2 + 1). (4)

If q divides 2(p−1)/2 − 1, then

q 6 2(p−1)/2 − 1, q + 1 < 2(p−1)/2 + 1.

Hence, we obtain

q(q + 1) < 2(2(p−1)/2 − 1)(2(p−1)/2 + 1),

a contradiction. If q divides 2(p−1)/2 +1, then 2(p−1)/2 +1 = kq. Now, from

Eq.( 4), it follows that

2k(2(p−1)/2 − 1) = q + 1.

When k = 1, we conclude that p = 5 and q = 5. But then, we have

|K/H| = |A2(5)| = 25 · 3 · 53 · 31,

while |G| = |C5(2)| = 225 · 36 · 52 · 7 · 11 · 17 · 31; this is a contradiction

because |K/H|5 > |G|5. If k > 2, then q > 2(2(p+1)/2 − 2)− 1. Therefore,

we have

2(2(p+1)/2 − 2)− 1 6 q < q + 1 6 2(p−1)/2 + 1,

a contradiction.

If (3, q − 1) = 3, then q(q + 1) = 22(3 · 2p−2 − 1), which implies that

(q + 1)2 = 4 and so (q − 1)2 = 2. Moreover, under these conditions, one

can easily deduce that f is odd, otherwise 8|q − 1 where

q − 1 = tf − 1 = (t f/2 − 1)(t f/2 + 1),

which is a contradiction. Thus, we have |A2(q)|2 = 24, while

|G/K|2 · |H|2 =|G|2|A2(q)|2

= 2p2−4.

Since |G/K| divides 2f(3, q − 1) and f is odd, |G/K|2 6 2. Therefore a

Sylow 2-subgroup of H has order either 2p2−4 or 2p

2−5. Applying Lemma

2.5 we deduce that 2p−1|2p2−4−1 or 2p−1|2p2−5−1. Now, one can easily

check that the second divisibility is possible only for p = 5. But then, we

get q(q + 1) = 22 · 23, which is a contradiction.

Step 3.30. K/H � 2Ar(q), where (q + 1)∣∣(r + 1) and (r, q) 6= (3, 3), (5, 2).


In this case, we have

|K/H| = |2Ar(q)| =1

(r + 1, q + 1)qr(r+1)/2

r+1∏i=2

(qi − (−1)i

),

and the only odd order component of 2Ar(q) is (qr + 1)/(q + 1). Therefore, we get

(qr + 1)/(q + 1) = 2p − 1.

An argument similar to that in the previous cases shows that

qr − 1 > (qr + 1)/(q + 1) = 2p − 1,

and so qr > 2p. Let q = tf , where t is a prime and f is a natural number. We now

consider three cases separately.

(i) r > 7. Then qr(r+1)/2 > q3(r+1) > 23q3r > 23(p+1), which forces by Lemma

2.9 that t = 2. Thus (2fr + 1)/(2f + 1) = 2p − 1, and, consequently, f = 1,

r = 3 and p = 2, which is a contradiction.

(ii) If r = 5, then (q5 + 1)/(q + 1) = 2p − 1. Arguing as in the case (i), we

conclude that t = 2 and f = 1, whence 12 = 2p, a contradiction.

(iii) If r = 3, then (q3+1)/(q+1) = 2p−1. It follows that q(q−1) = 2(2p−1−1),

and so q = p = 2, which is impossible.

Step 3.31. K/H � 2Ar−1(q).

In this case, we have

|K/H| = |2Ar−1(q)| = 1

(r, q + 1)qr(r−1)/2

r∏i=2

(qi − (−1)i

),

and the only odd order component of 2Ar−1(q) is (qr + 1)/(q + 1)(r, q + 1). Thus

qr + 1

(q + 1)(r, q + 1)= 2p − 1,

As before, we deduce that qr > 2p. Let q = tf , where t is a prime and f is a natural

number. We now consider three cases separately.

(i) r > 7. It follows that qr(r−1)/2 > q3r > 23p, which implies that t = 2 by

Lemma 2.9. Now, we obtain

2fr + 1

(2f + 1)(r, 2f + 1)= 2p − 1,

which contradicts Lemma 2.1 because 2p − 1 is the largest prime in π(G).


(ii) r = 5. In this case we have q5 + 1 = (q + 1)(2p − 1)(5, q + 1). Assume

first that q is even, that is q = 2f . If (5, q + 1) = 1, then we obtain

25f = 2f+p + 2p − 2f − 2, which is impossible. If (5, q + 1) = 5, then

25f = 5(2f+p + 2p − 2f )− 6, which is again a contradiction. Assume next

that q is odd. Noting that q(q−1)(q2+1) = (2p−1)(5, q+1)−1, it is easily

seen that the left hand side is congruent to 0 (mod 4), while the right hand

side is congruent to 2 (mod 4), a contradiction.

(iii) r = 3. In this case, we have (q3+1)/(q+1)(3, q+1) = 2p−1. If (3, q+1) = 1,

then we obtain

q(q − 1) = 2p − 2 = 2(2(p−1)/2 − 1)(2(p−1)/2 + 1).

If q divides 2, than p = 2, a contradiction. If q divides 2(p−1)/2 − 1 or

2(p−1)/2 + 1, then

q(q − 1) < 2p − 2 = 2(2(p−1)/2 − 1)(2(p−1)/2 + 1),

a contradiction. Therefore we may assume that (3, q + 1) = 3. If q is even,

then we conclude that q = 4, which is a contradiction. We now suppose

that q is odd. Since q(q− 1) = 22(3 · 2p−2− 1), it follows that (q− 1)2 = 4,

and so (q + 1)2 = 2. Moreover, under these hypotheses, one can easily

deduce that f is odd, otherwise 8|q − 1 = tf − 1 = (tf/2 − 1)(tf/2 + 1),

which is a contradiction. On the other hand, |G/K| divides f(3, q+ 1) and

since f is odd, |G/K|2 = 1. Therefore a Sylow 2-subgroup of H has order

2p2−4. Again, using Lemma 2.5, we see that 2p−1|2p2−4−1, which implies

that p = 2. This is a contradiction.

Step 3.32. K/H � Cr(2).

The only odd order component of Cr(2) is 2r − 1. Thus 2r − 1 = 2p − 1. It follows

that r = p, G/K = 1 and H = 1, which means G ∼= C. This completes the proof

of the theorem. �

Acknowledgment. We would like to thank Professor Arturo Magidin and the

referee for the careful reading of this manuscript and for many insightful suggestions

and remarks.

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M. Akbari and A. R. Moghaddamfar

Department of Mathematics,

K. N. Toosi University of Technology,

P. O. Box 16315-1618,

Tehran, Iran

e-mails: [email protected] (M. Akbari)

[email protected]; [email protected] (A. R. Moghaddamfar)

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